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Agata [3.3K]
3 years ago
7

What is the distance between points a and b (4,2) (9,5)

Mathematics
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer:

(5,3)

Step-by-step explanation:

first u have to lable x1,x2 and y1,y2 then subtract X2-X1 and Y2-Y1

which should look like this:

9-4=5x

5-2=3y

and now your distance should be (5,3) which is your x and y variable

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Find an equation of line with given slope and y-intercept. Slope= -6 y-intercept= (0,-10) The equation of the line is: ?
GREYUIT [131]

Step-by-step explanation:

so, we find the slope-intercept form :

y = ax + b

"a" is the slope (it is always the factor of x), "b" is the y-intercept (the y-value when x = 0).

we have the slope : -6

and the given point (0, -10) gives us already the y-value when x = 0 : -10

therefore, the equation is

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1 year ago
A pentagonal prism has a height of 9 feet, and each base edge is 3 feet. Find the volume. Round your answer to the nearest hundr
OLga [1]

Height of the pentagon = 9 ft

Edge of base = 3 ft

area of pentagon = \frac{s^{2}n}{4 tan\left ( \frac{180}{n} \right )}

s is the length of any side = 3

n is the number of sides = 5

tan is the tangent function calculated in degrees

\\ \\\\area of pentagon = \frac{3^{2}(5)}{4 tan\left ( \frac{180}{5} \right )}\\\\\\area of pentagon = \frac{9(5)}{4 tan\left ( 36 \right )}\\\\area of pentagon = \frac{45}{4  ( 0.7265 )}\\\\area of pentagon = \frac{45}{2.906}\\\\area of pentagon = 15.4852

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Volume of the pentagon = 15.4852 (9)

Volume of the pentagon = 139.366

Volume of the pentagon = 139.37 cu.ft


5 0
3 years ago
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Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,
Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

  • \displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}  }  

In order to use t*, we need to check conditions for using a t-distribution first.

  • Random for both samples -- NOT STATED in the problem ∴ <u><em>proceed with caution</em></u>!
  • Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
  • Normality: CLT is not met; <u>n < 30</u> for both locations A and B ∴ <u><em>proceed with caution</em></u>!

<u>Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.</u>

<u>Solve for t*:</u>

<u></u>

We need the <u>tail area </u>first.

  • \displaystyle \frac{1-.9}{2}= .05

Next we need the <u>degree of freedom</u>.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

  • df for A: 13 - 1 = 12
  • df for B: 18 - 1 = 17
  • df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding <u>t-score for df = 5 and tail area = .05</u>.

  • t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

  • \overline {x}_A=39
  • s_A=8
  • n_A=13
  • \overline {x}_B = 55
  • s_B=2
  • n_B=18
  • t^{*}=-2.015

Substitute the variables into the formula: \displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}  }.

  • 39-55 \  \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}

Simplify this expression.

  • -16 \ \pm \ -2.015 (\sqrt{5.1453} \ )
  • -16 \ \pm \ 3.73139

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

  • (-20.5707,-11.4293)

If we want to <u>interpret</u> the confidence interval of (-20.5707, -11.4293), we can say...

<u><em>We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.</em></u>

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a; x = y divided by two

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